A set of dice is intransitive (or nontransitive) if it contains n > 2 {\displaystyle n>2} dice, X 1 , X 2 , . . . , X n {\displaystyle X_{1},X_{2},...,X_{n}} with the property that X 1 {\displaystyle X_{1}} rolls higher than X 2 {\displaystyle X_{2}} more than half the time, X 2 {\displaystyle X_{2}} rolls higher than X 3 {\displaystyle X_{3}} more than half the time, and so on, but X 1 {\displaystyle X_{1}} does not roll higher than X n {\displaystyle X_{n}} more than half the time. In other words, a set of dice is intransitive if the binary relation – X rolls a higher number than Y more than half the time – on its elements is not transitive. More simply, X 1 {\displaystyle X_{1}} normally beats X 2 {\displaystyle X_{2}} , X 2 {\displaystyle X_{2}} normally beats X 3 {\displaystyle X_{3}} , but X 1 {\displaystyle X_{1}} does not normally beat X n {\displaystyle X_{n}} . It is possible to find sets of dice with the even stronger property that, for each die in the set, there is another die that rolls a higher number than it more than half the time. This is different in that instead of only " X 1 {\displaystyle X_{1}} does not normally beat X n {\displaystyle X_{n}} " it is now " X n {\displaystyle X_{n}} normally beats X 1 {\displaystyle X_{1}} ". Using such a set of dice, one can invent games which are biased in ways that people unused to intransitive dice might not expect (see example).
Example
Consider the following set of dice.
Die A has sides 2, 2, 4, 4, 9, 9. Die B has sides 1, 1, 6, 6, 8, 8. Die C has sides 3, 3, 5, 5, 7, 7. The probability that A rolls a higher number than B, the probability that B rolls higher than C, and the probability that C rolls higher than A are all 5/9, so this set of dice is intransitive. In fact, it has the even stronger property that, for each die in the set, there is another die that rolls a higher number than it more than half the time. Now, consider the following game, which is played with a set of dice.
The first player chooses a die from the set. The second player chooses one die from the remaining dice. Both players roll their die; the player who rolls the higher number wins. If this game is played with a set of transitive dice, it is either fair or biased in favor of the first player, because the first player can always find a die that will not be beaten by any other dice more than half the time. If it is played with the set of dice described above, however, the game is biased in favor of the second player, because the second player can always find a die that will beat the first player's die with probability 5/9. The following tables show all possible outcomes for all three pairs of dice.
If one allows weighted dice, i.e., with unequal probability weights for each side, then alternative sets of three dice can achieve even larger probabilities than 5 9 ≈ 0.56 {\displaystyle {\frac {5}{9}}\approx 0.56} that each die beats the next one in the cycle. The largest possible probability is one over the golden ratio, 1 φ ≈ 0.62 {\displaystyle {\frac {1}{\varphi }}\approx 0.62} .
Variations
Efron's dice Efron's dice are a set of four intransitive dice invented by Bradley Efron.
The four dice A, B, C, D have the following numbers on their six faces:
A: 4, 4, 4, 4, 0, 0 B: 3, 3, 3, 3, 3, 3 C: 6, 6, 2, 2, 2, 2 D: 5, 5, 5, 1, 1, 1 Each die is beaten by the previous die in the list with wraparound, with probability 2/3. C beats A with probability 5/9, and B and D have equal chances of beating the other. If each player has one set of Efron's dice, there is a continuum of optimal strategies for one player, in which they choose their die with the following probabilities, where 0 ≤ x ≤ 3/7:
P(choose A) = x P(choose B) = 1/2 - 5/6x P(choose C) = x P(choose D) = 1/2 - 7/6x
Miwin's dice
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